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Solve the inequality and graph the solution. – (n–3)>1 Plot the endpoints. Select an endpoint to change it from closed to open....Asked by evergreen
Solve the inequality and graph the solution.
q
2
≤3
Plot the endpoints. Select an endpoint to change it from closed to open. Select the middle of the segment, ray, or line to delete it.
q
2
≤3
Plot the endpoints. Select an endpoint to change it from closed to open. Select the middle of the segment, ray, or line to delete it.
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Answered by
GPT-4o mini
To solve the inequality \( q^2 \leq 3 \), we first rewrite it in a form that can be analyzed.
1. **Determine the roots**: Start by setting the inequality to zero:
\[
q^2 - 3 \leq 0
\]
This can be factored or solved by finding the square roots:
\[
q^2 = 3 \implies q = \pm \sqrt{3}
\]
Thus, the critical points are \( q = -\sqrt{3} \) and \( q = \sqrt{3} \).
2. **Test intervals**: The critical points divide the real number line into intervals. We need to test the inequality in the intervals \( (-\infty, -\sqrt{3}) \), \( (-\sqrt{3}, \sqrt{3}) \), and \( (\sqrt{3}, \infty) \).
- **Interval \( (-\infty, -\sqrt{3}) \)**: Choose \( q = -2 \)
\[
(-2)^2 = 4 \quad \text{(not }\leq 3\text{)}
\]
- **Interval \( (-\sqrt{3}, \sqrt{3}) \)**: Choose \( q = 0 \)
\[
0^2 = 0 \quad \text{(which is } \leq 3\text{)}
\]
- **Interval \( (\sqrt{3}, \infty) \)**: Choose \( q = 2 \)
\[
2^2 = 4 \quad \text{(not }\leq 3\text{)}
\]
3. **Conclusion**: The inequality is satisfied in the interval \( (-\sqrt{3}, \sqrt{3}) \) and at the endpoints.
The solution to the inequality is:
\[
[-\sqrt{3}, \sqrt{3}]
\]
4. **Graphing the solution**:
- Plot the points \( -\sqrt{3} \) and \( \sqrt{3} \):
- Since both endpoints are included, we will denote them with closed circles.
- The middle part between the two endpoints should be shaded.
5. **Adjusting the plot**: If we select an endpoint to change from closed to open, we could choose, for example, \( -\sqrt{3} \) to be open. The final graphical representation will look as follows:
- Plot:
- Open circle at \( -\sqrt{3} \) (not included)
- Closed circle at \( \sqrt{3} \) (included)
- Shade the interval between these two points.
This represents the solution to the inequality \( q^2 \leq 3 \) with the specified adjustments.
1. **Determine the roots**: Start by setting the inequality to zero:
\[
q^2 - 3 \leq 0
\]
This can be factored or solved by finding the square roots:
\[
q^2 = 3 \implies q = \pm \sqrt{3}
\]
Thus, the critical points are \( q = -\sqrt{3} \) and \( q = \sqrt{3} \).
2. **Test intervals**: The critical points divide the real number line into intervals. We need to test the inequality in the intervals \( (-\infty, -\sqrt{3}) \), \( (-\sqrt{3}, \sqrt{3}) \), and \( (\sqrt{3}, \infty) \).
- **Interval \( (-\infty, -\sqrt{3}) \)**: Choose \( q = -2 \)
\[
(-2)^2 = 4 \quad \text{(not }\leq 3\text{)}
\]
- **Interval \( (-\sqrt{3}, \sqrt{3}) \)**: Choose \( q = 0 \)
\[
0^2 = 0 \quad \text{(which is } \leq 3\text{)}
\]
- **Interval \( (\sqrt{3}, \infty) \)**: Choose \( q = 2 \)
\[
2^2 = 4 \quad \text{(not }\leq 3\text{)}
\]
3. **Conclusion**: The inequality is satisfied in the interval \( (-\sqrt{3}, \sqrt{3}) \) and at the endpoints.
The solution to the inequality is:
\[
[-\sqrt{3}, \sqrt{3}]
\]
4. **Graphing the solution**:
- Plot the points \( -\sqrt{3} \) and \( \sqrt{3} \):
- Since both endpoints are included, we will denote them with closed circles.
- The middle part between the two endpoints should be shaded.
5. **Adjusting the plot**: If we select an endpoint to change from closed to open, we could choose, for example, \( -\sqrt{3} \) to be open. The final graphical representation will look as follows:
- Plot:
- Open circle at \( -\sqrt{3} \) (not included)
- Closed circle at \( \sqrt{3} \) (included)
- Shade the interval between these two points.
This represents the solution to the inequality \( q^2 \leq 3 \) with the specified adjustments.
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